3-1 Solving Systems Using Tables and Graphs

Quick Review

A system of equations has two or more equations. Points of intersection are solutions. A linear system has linear equations. A consistent system can be dependent, with infinitely many solutions, or independent, with one solution. An inconsistent system has no solution.

Example

Solve the system { 3 x + 2 y = 4 2 x − 4 y = 8 left brace . table with 2 rows and 1 column , row1 column 1 , 3 x plus 2 y equals 4 , row2 column 1 , 2 x minus 4 y equals 8 , end table

Graph the equations.

The only solution, where the lines intersect, is ( 2 , − 1 ) . open 2 comma negative 1 close .

Exercises

Without graphing, classify each system of equations as independent, dependent, or inconsistent. Solve independent systems by graphing.

3. { 6 x − 2 y = 2 2 + 6 x = y left brace . table with 2 rows and 1 column , row1 column 1 , 6 x minus 2 y equals 2 , row2 column 1 , 2 plus 6 x equals y , end table
4. { 5 − y = 2 x 6 x − 15 = − 3 y left brace . table with 2 rows and 1 column , row1 column 1 , 5 minus y equals 2 x , row2 column 1 , 6 x minus 15 equals negative 3 y , end table
5. { 6 y + 2 x = 8 12 y + 4 x = 4 left brace . table with 2 rows and 1 column , row1 column 1 , 6 y plus 2 x equals 8 , row2 column 1 , 12 y plus 4 x equals 4 , end table
6. { 1.5 + 3 x = 0.5 y 6 − 2 y = − 12 x left brace . table with 2 rows and 1 column , row1 column 1 , 1.5 plus 3 x equals 0.5 y , row2 column 1 , 6 minus 2 y equals negative 12 x , end table
7. { 2 − 0.25 x = 0.5 y − 1.5 y = 1.5 x − 3   left brace . table with 3 rows and 1 column , row1 column 1 , 2 minus , 0.25 , x equals 0.5 y , row2 column 1 , negative 1.5 y equals 1.5 x minus 3 , row3 column 1 , , end table
8. { 1 + y = x x + y = 1 left brace . table with 2 rows and 1 column , row1 column 1 , 1 plus y equals x , row2 column 1 , x plus y equals 1 , end table
9. For $7.52, you purchased 8 pens and highlighters from a local bookstore. Each highlighter cost $1.09 and each pen cost $.69. How many pens did you buy?

3-2 Solving Systems Algebraically

Quick Review

To solve an independent system by substitution, solve one equation for a variable. Then substitute that expression into the other equation and solve for the remaining variable. To solve by elimination, add two equations with additive inverses as coefficients to eliminate one variable and solve for the other. In both cases you solve for one of the variables and use substitution to solve for the remaining variable.

Example

Solve the system { 10 − y = 4 x x = 4 + 0.5 y left brace . table with 2 rows and 1 column , row1 column 1 , 10 minus y equals 4 x , row2 column 1 , x equals 4 plus 0.5 y , end table  by substitution.

Substitute for x: 10 − y = 4 ( 4 + 0 . 5 y ) = 16 + 2 y . 10 minus y equals 4 open 4 plus 0 . 5 y close equals 16 plus 2 y .

Solve for y: y = − 2 . y equals negative 2 .

Substitute into the first equation:

10 − ( − 2 ) = 4 x . 10 minus open negative 2 close equals 4 x .

Solve for x: x = 3. The solution is ( 3 , − 2 ) . open 3 comma negative 2 close .

Exercises

Solve each system by substitution.

10. { x − 2 y = 3 3 x + y = − 5 left brace . table with 2 rows and 1 column , row1 column 1 , x minus 2 y equals 3 , row2 column 1 , 3 x plus y equals negative 5 , end table
11. { 14 x − 35 = 7 y − 25 − 6 x = 5 y left brace . table with 2 rows and 1 column , row1 column 1 , 14 x minus 35 equals 7 y , row2 column 1 , negative 25 minus 6 x equals 5 y , end table

Solve each system by elimination.

12. { 11 − 5 y = 2 x 5 y + 3 = − 9 x left brace . table with 2 rows and 1 column , row1 column 1 , 11 minus 5 y equals 2 x , row2 column 1 , 5 y plus 3 equals negative 9 x , end table
13. { 2 x + 3 y = 4 4 x + 6 y = 9 left brace . table with 2 rows and 1 column , row1 column 1 , 2 x plus 3 y equals 4 , row2 column 1 , 4 x plus 6 y equals 9 , end table
14. Roast beef has 25 g of protein and 11 g of calcium per serving. A serving of mashed potatoes has 2 g of protein and 25 g of calcium. How many servings of each are needed to supply exactly 29 g of protein and 61 g of calcium?

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